Let $a$,$b$,and $c$ be $3$ non-zero vectors such that no $2$ of these are collinear. If vector $a + 2b$ is collinear with $c$ and $b + 3c$ is collinear with $a$,then $a + 2b + 6c$ equals:

Let $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}$,$\vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$ and $\vec{c}$ be vectors such that $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. If $\vec{a} \cdot \vec{c}=-12$ and $\vec{c} \cdot (\hat{i}-2 \hat{j}+\hat{k})=5$,then $\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k})$ is equal to $.............$.

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are three vectors such that $\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}$ and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is $\frac{\pi}{2}$,then:

If $a, b, c$ are coplanar vectors,then

Area of a rectangle having vertices $A, B, C$ and $D$ with position vectors $-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$,$\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$,$\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$ and $-\hat{i} - \frac{1}{2}\hat{j} + 4\hat{k}$,respectively is . . . . . . .

The angle between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ is